On Tue, Oct 26, 2021 at 12:14 PM Steven Rostedt <rostedt@xxxxxxxxxxx> wrote:
>
> On Mon, 25 Oct 2021 13:08:38 -0700
> Kalesh Singh <kaleshsingh@xxxxxxxxxx> wrote:
>
> > == Results ==
> >
> > Divisor is a power of 2 (divisor == 32):
> >
> > test_hist_field_div_not_optimized | 8,717,091 cpu-cycles
> > test_hist_field_div_optimized | 1,643,137 cpu-cycles
> >
> > If the divisor is a power of 2, the optimized version is ~5.3x faster.
> >
> > Divisor is not a power of 2 (divisor == 33):
> >
> > test_hist_field_div_not_optimized | 4,444,324 cpu-cycles
> > test_hist_field_div_optimized | 5,497,958 cpu-cycles
>
> To optimize this even more, if the divisor is constant, we could make a
> separate function to not do the branch, and just shift or divide.
Ack. I can update to use separate functions for the constant divisors.
>
> And even if it is not a power of 2, for constants, we could implement a
> multiplication and shift, and guarantee an accuracy up to a defined max.
>
>
> If div is a constant, then we can calculate the mult and shift, and max
> dividend. Let's use 20 for shift.
>
> // This works best for small divisors
> if (div > max_div) {
> // only do a real division
> return;
> }
> shift = 20;
> mult = ((1 << shift) + div - 1) / div;
> delta = mult * div - (1 << shift);
> if (!delta) {
> /* div is a power of 2 */
> max = -1;
> return;
> }
> max = (1 << shift) / delta;
I'm still trying to digest the above algorithm. But doesn't this add 2
extra divisions? What am I missing here?
Thanks,
Kalesh
>
> We would of course need to use 64 bit operations (maybe only do this for 64
> bit machines). And perhaps even use bigger shift values to get a bigger max.
>
> Then we could do:
>
> if (val1 < max)
> return (val1 * mult) >> shift;
>
> -- Steve
